Find the equation of the plane through the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 which is perpendicular to the plane 5x + 3y – 6z + 8 = 0?
we know that, equation of a plane passing through the line of intersection of two planes
a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given by
(a1x + b1y + c1z + d1) + k(a2x + b2y + c2z + d2) = 0
So equation of plane passing through the line of intersection of given two planes
x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 is given by,
(x + 2y + 3z – 4) + k(2x + y – z + 5) = 0
x + 2y + 3z – 4 + 2kx + ky – kz + k5 = 0
x(1 + 2k) + y(2 + k) + z(3 – k) – 4 + 5k = 0 …… (1)
we know that, two planes are perpendicular if
a1a2 + b1b2 + c1c2 = 0 …… (2)
given, plane (1) is perpendicular to plane,
5x + 3y – 6z + 8 = 0 …… (3)
Using (1) and (3) in equation (2)
(5)(1 + 2k) + (3)(2 + k) + ( – 6)(3 – k) = 0
5 + 10k + 6 + 3k – 18 + 6k = 0
– 7 + 19k = 0
put the value of k in equation (1)
x(1 + 2k) + y(2 + k) + z(3 – k) – 4 + 5k = 0
x(1 + 
) + y(2 + 
) + z(3 – 
) – 4 + 
= 0
x(
) + y(
) + z(
) + 
= 0
x(
) + y(
) + z(
) – 
= 0
multiplying with 19 we get
33x + 45y + 50z – 41 = 0
Equation of required plane is, 33x + 45y + 50z – 41 = 0